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将列表分成N个长度大致相等的部分

更新时间:2022-04-25 21:34:57

由于四舍五入错误,该代码已损坏.不要使用它!

assert len(chunkIt([1,2,3], 10)) == 10  # fails


这里是可行的:


Here's one that could work:

def chunkIt(seq, num):
    avg = len(seq) / float(num)
    out = []
    last = 0.0

    while last < len(seq):
        out.append(seq[int(last):int(last + avg)])
        last += avg

    return out

测试:

>>> chunkIt(range(10), 3)
[[0, 1, 2], [3, 4, 5], [6, 7, 8, 9]]
>>> chunkIt(range(11), 3)
[[0, 1, 2], [3, 4, 5, 6], [7, 8, 9, 10]]
>>> chunkIt(range(12), 3)
[[0, 1, 2, 3], [4, 5, 6, 7], [8, 9, 10, 11]]